Question

Prove that if 3n^2+5 is odd, then n is even. Please post a detailed answer. Thank you.

Answer 1

first, you need to show that only

even+odd = odd

even+even = 2m+2n = 2(m+n) = even
odd+odd = 2m+1 + 2n+1 = 2m+2n+2 = even
even+odd = 2m + 2n+1 = 2(m+n)+1 is odd

so, if 3n^2+5 is odd, 3n^2 is even
Since 3 is odd, n^2 must be even, which means that n is even, since

odd*odd = (2m+1)(2m+1) = 4m^2+4m+1 = 4(m^2+m) + 1 is odd